Question

factorise it be fast please

Answer 1

Hello Friends ✌
Here is u r Ans ➡

✨
$= \frac{2x {}^{2} - x - 3}{3x {}^{2} + 5x + 2}$

$= \frac{2x {}^{2} + 2x - 3x - 3 }{3x {}^{2} + 3x + 2x + 2}$

$= \frac{2x \times (x + 1) - 3x - 3}{3x {}^{2} + 3x + 2x + 2 }$

$= \frac{2x \times (x + 1) - 3(x + 1)}{3x \times (x + 1) + 2(x + 1)}$

$= \frac{(x + 1) \times (2x - 3)}{(x + 1) \times (3x + 2)}$

$= \frac{2x - 3}{3x + 2}$

✨b) ➡
$= 16p {}^{2} + \frac{8}{3} pq + \frac{1}{9} q {}^{2}$


$= using \: > \: a {}^{2} + 2ab + b {}^{2} = (a + b) {}^{2}$

$= (4p + \frac{1}{3} q) {}^{2}$


✨c) ➡
$= 121t {}^{2} - 0.04$

$= 121t {}^{2} - \frac{1}{25}$

$= using \: > \: a {}^{2} - b {}^{2} = (a - b)(a + b)$

$= (11t - \frac{1}{5} ) \times (11t + \frac{1}{5} )$
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Answer 2

(a)

Given Equation is (2x^2 - x - 3)/(3x^2 + 5x + 2)

⇒ (2x^2 - 3x + 2x - 3)/(3x^2 + 3x + 2x + 2)

⇒ x(2x - 3) + 1(2x - 3)/3x(x + 1) + 2(x + 1)

⇒ (x + 1)(2x - 3)/(3x + 2)(x + 1).

⇒ (2x - 3)/(3x + 2)

(b)

Given Equation is 16p^2 + (8/3) pq + (1/9)q^2

⇒ (1/9) [144p^2 + (72/3) pq + q^2]

⇒ (1/9)[(12p)^2 + 24 pq + q^2]

⇒ (1/9)[12p + q]^2.

(or)

⇒ (4p)^2 + 2(4)(1/3) + (1/3q)^2

⇒ (4p + 1/3q)^2

(c)

Given Equation is 121t^2 - 0.04

⇒ (11t)^2 - (0.2)^2.

We know that a^2 - b^2 = (a + b)(a - b)

⇒ (11t + 0.2)(11t - 0.2).

Hope it helps!